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Answer:
For a 95% confidence interval, the veterinarian should choose the critical value to be 2.093.
Step-by-step explanation:
We are given that a veterinarian collects data on the number of times race horses are raced during their careers. The veterinarian finds that the average number of races a horse enters is [tex]\bar x[/tex] = 15.3 , with a standard deviation of s = 6.8 in a sample of n = 20 horses.
Now, we have to find that what critical value would he choose for a 95% confidence interval.
Firstly, Confidence interval indicate that how much confident we are that our true parameter value will be in the required interval estimated.
And critical value tells that it is that value in the t table whose less than area is 95% (for 5% significance level).
Now, here we look our critical value in t table because we are not known with population standard deviation ([tex]\sigma[/tex]). And also, we assume the test to be two-tailed.
So, in the t table the critical value of t at 19 (n-1) degree of freedom at 2.5% significance level (two-tailed) is given to be 2.093.
Therefore, for a 95% confidence interval, the veterinarian should choose the critical value to be 2.093.
The veterinarian should choose the critical value to be "2.093".
Given:
Sample mean,
- [tex]\bar x = 15.3[/tex]
Standard deviation,
- [tex]s = 6.8[/tex]
Sample size,
- [tex]n = 20[/tex]
Degrees of freedom will be:
[tex]df=n-1[/tex]
[tex]= 20-1[/tex]
[tex]=19[/tex]
At 95%, the confidence level the "t" will be:
→ [tex]\alpha = 1-95[/tex]%
[tex]= 1-0.95[/tex]
[tex]= 0.05[/tex]
→ [tex]\frac{\alpha}{2} = \frac{0.05}{2}[/tex]
[tex]= 0.025[/tex]
hence,
= [tex]t_{\frac{\alpha}{2} }[/tex]
= [tex]t_{0.025,19}[/tex]
= [tex]2.093[/tex]
Thus the response above is correct.
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